Question: What is the slope of the line tangent to $f(x) = -2x^{2}+3x+4$ at $x = 1$ ?
The slope of the tangent line is $ \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}$ $ = \lim_{\Delta x \to 0} \frac{(-2(x+\Delta x)^{2}+3(x+\Delta x)+4) - (-2x^{2}+3x+4)}{\Delta x}$ $ = \lim_{\Delta x \to 0} \frac{(-2(x^{2}+2x \Delta x+\Delta x^{2})+3(x+\Delta x)+4) - (-2x^{2}+3x+4)}{\Delta x}$ $ = \lim_{\Delta x \to 0} \frac{-2x^{2}-4(x \Delta x)-2\Delta x^{2}+3x+3(\Delta x)+4+2x^{2}-3x-4}{\Delta x}$ $ = \lim_{\Delta x \to 0} \frac{-4(x \Delta x)-2\Delta x^{2}+3(\Delta x)}{\Delta x}$ $ = \lim_{\Delta x \to 0} -4x-2(\Delta x)+3$ $ = -4x+3$ $ = (-4)(1)+3$ $ = -1$